CHAPTER 10 Principles of Chemical Thermodynamics and · PDF fileenergy transfer, in which...

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CHAPTER 10

Principles of ChemicalThermodynamics

and Kinetics

Read This Chapter to Learn About➤ Enzymes

➤ Principles of Bioenergetics

➤ Thermodynamics: Energy Changes in Chemical Reactions

➤ Kinetics and Equilibrium: Rate Processes in Chemical Reactions

ENZYMESThe cell, the basic unit of life, acts as a biochemical factory, using food to produce

energy for all the functions of life, including growth, repair, and reproduction. This

work of the cell involves many complex chemical processes, including respiration and

energy transfer, in which specific enzymes are used to facilitate the reactions.

Classification of EnzymesEnzymes are a special category of proteins that serve as biological catalysts, speeding

up chemical reactions. The enzymes, with names often ending in the suffix –ase, are

essential to the maintenance of homeostasis, or a stable internal environment, within

a cell. The maintenance of a stable cellular environment and the functioning of the cell

are essential to life.

337


338UNIT II:ChemicalFoundations ofBiological Systems

(a) (b)

Reactant Reactant

Product

Ener

gy s

uppl

ied

Ener

gy re

leas

ed

Product

Activationenergy

Activationenergy

Catalyzed

Uncatalyzed

FIGURE 10-1 Lowering activation energy. (a) Activation energy is the amount of energy neededto destabilize chemical bonds. (b) Enzymes serve as catalysts to lower the amount of activationneeded to initiate a chemical reaction. Source: From George B. Johnson, The Living World, 3rded., McGraw-Hill, 2003; reproduced with permission of The McGraw-Hill Companies.

Enzymes function by lowering the activation energy (see Figure 10-1) required to

initiate a chemical reaction, thereby increasing the rate at which the reaction occurs.

Enzymes are involved in catabolic reactions that break down molecules, as well as

in anabolic reactions that are involved in biosynthesis. Most enzymatic reactions are

reversible. Enzymes are unchanged during a reaction and are recycled and reused.

Enzyme StructureAs stated earlier, enzymes are proteins and, like all proteins, are made up of amino

acids. Interactions between the component amino acids determine the overall shape

of an enzyme, and it is this shape that is critical to an enzyme’s ability to catalyze a

reaction.

The area on an enzyme where it interacts with another substance, called a sub-

strate, is the enzyme’s active site. Based on its shape, a single enzyme typically only

interacts with a single substrate (or single class of substrates); this is known as the

enzyme’s specificity. Any changes to the shape of the active site render the enzyme

unable to function. Enzyme kinetics or the rate of biochemical reactions is described

by the Michaelis–Menten mechanism:

[enzyme] + [substrate]k1

k−1

[enzyme − substrate complex]k2⇒ enzyme + products

where k1, k−1, and k2 are reaction rate constants.


339CHAPTER 10:Principles of

ChemicalThermodynamics

and Kinetics

Enzyme FunctionThe induced fit model is used to explain the mechanism of action for enzyme func-

tion seen in Figure 10-2. Once a substrate binds loosely to the active site of an enzyme,

a conformational change in shape occurs to cause tight binding between the enzyme

and the substrate. This tight binding allows the enzyme to facilitate the reaction. A sub-

strate with the wrong shape cannot initiate the conformational change in the enzyme

necessary to catalyze the reaction.

Enzyme–substrate complex

FIGURE 10-2 The induced fit model. Enzymes interact with their substrates to form an enzyme–substrate complex. This complex allows the chemical reaction to occur. Source: From GeorgeB. Johnson, The Living World, 3rd ed., McGraw-Hill, 2003; reproduced with permission of TheMcGraw-Hill Companies.

Some enzymes require assistance from other substances to work properly. If assis-

tance is needed, the enzyme has binding sites for cofactors or coenzymes. Cofactors

are various types of ions such as iron and zinc (Fe2+ and Zn2+). Coenzymes are organic

molecules usually derived from vitamins obtained in the diet. For this reason, mineral

and vitamin deficiencies can have serious consequences on enzymatic functions.

Factors That Affect Enzyme FunctionThere are several factors that can influence the activity of a particular enzyme. The

first is the concentration of the substrate and the concentration of the enzyme.

Reaction rates stay low when the concentration of the substrate is low, whereas the

rates increase when the concentration of the substrate increases. Temperature is also

a factor that can alter enzyme activity. Each enzyme has an optimal temperature for

functioning. In humans, this is typically body temperature (37 ◦C). At lower temper-

atures, the enzyme is less efficient. Increasing the temperature beyond the optimal

point can lead to enzyme denaturation, which renders the enzyme useless. Enzymes



also have an optimal pH in which they function best, typically around 7 in humans,

although there are exceptions. Extreme changes in pH can also lead to enzyme denat-

uration. The denaturation of an enzyme is not always reversible.

Control of Enzyme ActivityIt is critical to be able to regulate the activity of enzymes in cells to maintain efficiency.

This regulation can be carried out in several ways. Feedback, or allosteric inhibition,

illustrated in Figure 10-3, acts somewhat like a thermostat to regulate enzyme activity.

Many enzymes contain allosteric binding sites and require signal molecules such as

repressors and activators to function. As the product of a reaction builds up, repressor

molecules can bind to the allosteric site of the enzyme, causing a change in the shape

of the active site. The consequence of this binding is that the substrate can no longer

interact with the active site of the enzyme and the activity of the enzyme is temporar-

ily slowed or halted. When the product of the reaction declines, the repressor molecule

dissociates from the allosteric site. This allows the active site of the enzyme to resume

its normal shape and normal activity. Some allosteric enzymes stay inactive unless

activator molecules are present to allow the active site to function.

FIGURE 10-3 Allosteric inhibition of an enzyme. Repressors can be used to regulate the activityof an enzyme. Source: From George B. Johnson, The Living World, 3rd ed., McGraw-Hill, 2003;reproduced with permission of The McGraw-Hill Companies.

Inhibitor molecules also regulate enzyme action. A competitive inhibitor is a

molecule that resembles the substrate in shape so much that it binds to the active site

of the enzyme, thus preventing the substrate from binding. This halts the activity of the

enzyme until the competitive inhibitor is removed or is outcompeted by an increasing

amount of substrate. Noncompetitive inhibitors bind to allosteric sites and change

the shape of the active site, thereby decreasing the functioning of the enzyme. Increas-

ing levels of substrate have no effect on noncompetitive inhibitors, but the activity of

the enzyme can be restored when the noncompetitive inhibitor is removed.




and Kinetics

PRINCIPLES OF BIOENERGETICS

Overview of MetabolismLiving organisms maintain their systems in a dynamic steady state by taking in food.

Energy is extracted from food to build complex molecules from simpler ones, and for

storage. Collectively, these processes are called metabolism, the enzyme-catalyzed

transformation of energy and matter.

The metabolic pathways are a sequence of enzymatic reactions that serves to trans-

form energy and matter. In these sequences, the product of one reaction is the sub-

strate for the next. Intermediates are called metabolites. There are two aspects of

metabolism: catabolism and anabolism. Catabolism involves the breakdown of com-

plex molecules into simpler ones. Food molecules are broken down to building block

molecules and energy. Some of this energy is stored as ATP (adenosine triphosphate).

Anabolism involves the biosynthesis of complex molecules from the building blocks.

The energy needed for this comes from stored ATP, from high-energy hydrogen in the

form of NADPH (nicotinamide adenine dinucleotide phosphate).

Bioenergetics/ThermodynamicsMetabolism is quantified using principles of thermodynamics: enthalpy, entropy, and

free energy. It is the �H, �S, and �G that is considered. Enthalpy is the heat content.

The �H of a reaction can be determined from the sum of the �H f of the products

minus the �Hf of the reactants. The entropy of a system is the degree of random-

ness. The more random the system, the higher the S. S cannot be measured, but can be

obtained from �G = �H−T�S where �G, �H, and T can be measured. The free energy

is the amount of useful work that can be obtained from a system at constant temper-

ature, pressure, and volume. The change in free energy �G◦ can be derived from the

equilibrium constant K of a reaction, where �G◦ = −RT ln K . R is the gas constant

8.314 J/mol K and T is temperature in Kelvin. �G is a measure of the spontaneity of a

reaction. If it is a negative value, the reaction is spontaneous (exothermic), and if it is

a positive value, the reaction is endothermic.

Phosphorylation/ATPAn unfavorable reaction (+�G) can be driven forward by coupling it to a favorable one,

so that the net �G is negative. For instance, the formation of glucose-6-phosphate from

glucose and phosphate (Pi) is unfavorable, with a �G of +3.3 kcal. If this reaction is

coupled to the hydrolysis of ATP to ADP and Pi , which has a �G = −7.3 kcal, then

glucose can be phosphorylated with a net �G = −4.0 kcal.



Phosphorylation Reaction ΔG

ATP + H2O −→ ADP + Pi −7.3 kcal

Glucose + Pi −→ Glucose-6-phosphate +3.3 kcal

Glucose + ATP + H2O −→ Glucose-6-phosphate + ADP −4.0 kcal

ATP is a high-energy phosphate compound. As such, it has a highly negative �G

for the hydrolysis of phosphate. All high-energy phosphate compounds have excel-

lent phosphate group transfer potential. ATP transfers phosphate to many sugars and

glycerol. The enzymes that catalyze this process are called hexokinases and glycerol

kinases. Kinases are transfer enzymes. All high-energy phosphate compounds pass Pi

to lower energy acceptors via ATP. The �G of ATP hydrolysis, at −7.3 kcal, is interme-

diate between compounds of high phosphate transfer potential and those with lower

phosphate transfer potential. ATP functions as the carrier of phosphate between the

catabolic pathways and the anabolic pathways.

Sometimes two terminal phosphates of ATP are removed as pyrophosphate, PPi ,

leaving AMP. Energy is obtained by the hydrolysis of pyrophosphate to two phosphates

by the enzyme pyrophosphatase as shown:

PPi + H2O −→ 2Pi �G = −7.2 kcal

The AMP reacts with an ATP to form two ADPs by the enzyme adenylate kinase in

the reaction:

AMP + ATP −→ 2ADP

Other nucleoside triphosphates are used as well. GTP is used in protein synthesis,

CTP is used in lipid synthesis, and UTP is used in polysaccharide synthesis. The NDP

or NMP formed is re-phosphorylated by ATP in the reaction:

UDP + ATP −→ UTP + ADP

High-energy phosphate compounds act as storage forms of energy. The turnover of

ATP is very high. One’s body weight in ATP is formed and broken down every 24 hours.

Therefore, ATP is not the greatest for energy storage. Creatine phosphate is the storage

form of phosphate in humans. During rest, creatine is re-phosphorylated by ATP.

Oxidation–Reduction in Biological SystemsRedox reactions are electron-transfer reactions. Oxidation is the loss of electrons while

reduction is the gain of electrons. The compounds NAD and NADPH are the common

reducing agents in metabolic pathways.

NADH −→ NAD+ + H+ + 2e−

NADPH −→ NADP+ + H+ + 2e−




and Kinetics

The electrons given off are ultimately used to reduce oxygen to water. The ultimate

acceptor of electrons derived from food molecules is oxygen. Food molecules transfer

electrons to electron carriers such as NAD+ or FAD. The electrons from the carriers

reach oxygen via the electron transport system. Both NAD+ and FAD are the cofactors

for the dehydrogenase enzymes involved in oxidation of food molecules. The break-

down pathways are oxidative, while the anabolic pathways are reductive and use

NADPH as the electron donor.

THERMODYNAMICS: ENERGY CHANGES INCHEMICAL REACTIONSThermodynamics is the study of the relationships of enthalpy, entropy (disorder of a

system), and free energy. The laws of thermodynamics and certain formulas explain

the relationships between free energy, entropy, work, temperature, and equilibrium.

Thermodynamic System and State FunctionsA state function is a property of a system that depends upon its current condition

only. It will not matter how one arrives at that condition. These functions could

include �E (energy change), �U (internal energy change), �H (enthalpy change), �S

(entropy change), or �G (Gibbs free energy change). Examples that are not state func-

tions are work (W) and heat (Q).

Zeroth Law of ThermodynamicsThe zeroth law of thermodynamics explains what happens when three systems are in

thermal equilibrium with each other. To summarize, should two systems be separate

from each other but both be in thermal equilibrium with a third, then the first two are

also in thermal equilibrium with each other. For example, if in a closed system you have

three solid metal cubes all at different temperatures and in contact with each other,

the heat will flow from the high temperatures to the low temperatures until a thermal

equilibrium is reached.

First Law of ThermodynamicsThe first law of thermodynamics states that the change in the internal energy of a

system is given by �U = Q + W, where Q = heat and W = work. U is the sum of the

kinetic and potential energies. It is a state function; that is, it depends only on the initial

and final states, and not on the path between. The heat Q is the energy into or out of a



system because of a temperature difference between the system and its surroundings.

Work W is the energy that results when a force moves an object some distance, d.

The enthalpy H is equal to the quantity U+PV , where H is also a state function. �H

is the difference between the initial and final states; at atmospheric pressure, �H = Q.

Second Law of ThermodynamicsThe second law of thermodynamics states that the total entropy of a system and its

surroundings always increases for a spontaneous process. This law relates spontaneity

to entropy. The effect on the surroundings must be taken into account.

For a spontaneous process, �S must be >Q/T , and at equilibrium, �S = Q/T =�H/T .

ENTROPY: A MEASURE OF SYSTEM DISORDER

Entropy, represented by the symbol S, is a measure of the disorder of a system. S is

also a state function, and �S is the difference between the initial and final states.

If �S is positive, the process results in more disorder. If �S is negative, the process

results in more order.

If disorder is increasing, �Srxn is positive. Some examples of processes with

increasing disorder include more gas produced than used up; a solid converted to a

liquid; and a solid dissolved in a solvent.

If order is increasing, �Srxn is negative. Some examples of processes that have

increasing order include less gas produced than used up; a liquid converted to a solid;

and a solid precipitated from a solution.

RELATIVE ENTROPY FOR GAS, LIQUID, AND SOLID STATES

Different phases of matter will have different degrees of freedom and randomness. The

more disorder present in a system, the more entropy that system will have. The letter

symbol for entropy is S, and because it is a state function, you look at only the final and

initial entropy of a system, �S.

Because gas molecules are spread out and have a continuous random, straight-line

motion with (ideally) no forces of attraction between them, the entropy of gases is far

greater than that of liquids. Liquids and solids both have their molecules touching, but

the molecules in liquids have more freedom than those of solids. To summarize �S for

the phases: gas >> liquid > solid.

Examples of processes in which �S is negative might be raking up leaves and plac-

ing them in a bag or condensing steam into a liquid. Examples of processes in which �S

is positive might be the wind blowing those same leaves all over or you spilling a glass

of water. One important concept to keep in mind is that the entropy of the universe




and Kinetics

is positive. All of the matter in the universe is believed to have been together at one

time until the Big Bang occurred. Since then, the universe has been expanding, which

explains why processes in the universe favor a positive value for �S. Think about the

last time you dropped an egg by accident. When was the last time you saw the egg

white, yolk, and shell come back together on its own to re-form the egg?

CalorimetryHEAT

Heat is closely associated with temperature, but they are very different quantities. Heat,

or thermal energy, is a form of energy that depends on a change of temperature and

can be converted to work and other forms of energy. The quantity of heat Q gained or

lost by a body is related to the temperature difference �T by the following equation:

Q = mc�T

where m is the mass of the body and c is a proportionality constant termed the

specific heat capacity. Specific heat capacity is the heat needed to raise the temper-

ature of 1 g of a substance by 1◦C. It has the symbol c and the units J/g ◦C. Thus,

Q = mc�T , where �T is the difference in the initial and final temperatures. The fol-

lowing table lists the specific heats of some common substances.

TABLE 10-1 The Specific Heat of Some Common Substances

Substance Specific Heat in J/g ◦C

Al 0.900Au 0.129C (diamond) 0.502Cu 0.385Fe 0.444Hg 0.139H2O 4.184

EXAMPLE: Calculate the heat needed to raise the temperature of 85.3 g of iron

from 35.0 ◦C to 275.0 ◦C.

SOLUTION:

➤ The �T is 275.0 ◦C − 35.0 ◦C = 240.0 ◦C.

➤ The c for iron is 0.444 J/g ◦C, given from the table.

➤ Plug the values into the equation and solve for Q :

Q = mc�T = (0.444 J/g ◦C) (85.3 g) (240.0 ◦C) = 9.09 × 103 J



Calorimetry involves the use of an insulated container at room pressure to mea-

sure the heat changes that occur during a physical or chemical process. Because the

calorimeter is insulated, no heat can leave or enter the system, and all of the heat

changes that occur within the calorimeter are zero. In other words, the heat that is

given off equals the heat that is gained.

Heat released = heat absorbed

and �T = higher temperature − lower temperature

The calorimeter itself is ignored in the following calculations.

EXAMPLE: Calculate the final temperature of the system when an 18.4-g sam-

ple of aluminum at 215 ◦C is added to a calorimeter that contains 120.0 g water

at 24.5 ◦C.

SOLUTION:

➤ The c for water is 4.184 J/g ◦C.

➤ The c for aluminum is 0.900 J/g ◦C, given from the table.

➤ The �T for the water is x − 24.5 ◦C, where x is the final temperature. For

water, x is the higher temperature.

➤ The �T for the aluminum is 215 ◦C − x, where x is the final temperature. For

the aluminum, 215 ◦C is the higher temperature.

➤ Plugging into the equation (mc�T)H2O = (mc�T )Al and solving for x :

(120.0 g)(4.184 J/g ◦C)(x − 24.5 ◦C) = (18.4 g) (0.900 J/g ◦C)(215 ◦C − x)

x = 30.6 ◦C

Heat of transformation is similar to the specific heat capacity, but it accounts for

changes in the phase of the body. The specific heat of a body assumes that no change in

phase occurs during a temperature change. In order for a substance to change states of

matter—that is, from solid to liquid or from liquid to gas—heat energy must be added

to or removed from the substance. The amount of heat required to change the phase

of 1 kg of a substance is the heat of transformation L. Thus, the total amount of heat Q

gained or lost by a substance of mass m during a change between phases is:

Q = mL

where L is the heat of transformation unique to the substance. The heat of transfor-

mation exists in two forms, according to the particular phase transformation:

➤ Heat of fusion. L f is the amount of heat energy required to change 1 kg of solid

matter to liquid or the amount of energy released when changing 1 kg of liquid

matter to solid.

➤ Heat of vaporization. Lv is the amount of heat energy required to change 1 kg of

liquid matter to gas or the amount of energy released when changing 1 kg of gas

matter to liquid.




and Kinetics

Heat TransferThere are three mechanisms of heat transfer: conduction, convection, and radiation,

each dependent on the state of matter of the object.

Conduction is the method of heat energy transfer that occurs in solids—for exam-

ple, the warming of a spoon when cream is stirred in coffee. In conduction, heat energy

is transferred by collisions between the rapidly moving molecules of the hot region and

the slower-moving molecules of the cooler region. A portion of the kinetic energy from

the rapidly moving molecules is transferred to the slower-moving molecules, causing

an increase in heat energy at the cooler end and a subsequent increase in the flow of

heat.

Convection is the method of heat energy transfer that occurs in liquids and

gases—for example, the cooling of coffee after cream is poured into it. Convection rep-

resents the transfer of heat energy due to the physical motion or flow of the heated

substance, carrying heat energy with it to cooler regions of the substance. In contrast

to conduction, convection is the primary mechanism of heat transfer in fluids.

Radiation is the method of heat energy transfer that occurs in space, for exam-

ple, Earth’s surface being warmed by the sun. Radiation represents the transfer of heat

energy by electromagnetic waves that are emitted by rapidly vibrating, electrically-

charged particles. The electromagnetic waves propagate from the heated body or

source at the speed of light.

Endothermic and Exothermic ReactionsHeat is an energy that flows into or out of a system due to a difference between the

temperature of the system and its surroundings when they are in thermal contact. Heat

flows from hotter to cooler areas and has the symbol Q. If heat is added to a system, Q

is positive. If heat is removed from a system, Q is negative. The units of heat are joules

(J) or calories. One calorie equals 4.184 J.

A reaction from which heat is given off and has a negative Q is called an exother-

mic reaction. A reaction that requires heat to be added and has a positive Q is called

an endothermic reaction.

ENTHALPY AND STANDARD HEATS OF REACTION AND FORMATION

Enthalpy is a state function and it is used to denote heat changes in a chemical reac-

tion. It is given the symbol H. Because it depends only on the initial and final states

of the system, the difference between these states is �H. At standard pressure (1 atm),

Q = �H.

�H◦f is called the enthalpy of formation of a substance. It denotes the heat that

is absorbed or given off when a substance is produced from its elements at standard

temperature (25 ◦C) and pressure (1 atm).



�H◦rxn is called the enthalpy of reaction. It denotes the amount of heat that is given

off (−�H) or absorbed (+�H) by a reaction at standard temperature and pressure.

The �H◦rxn can be calculated from �H◦

f values. For the reaction

aA + bB → cC + dD,

the �H◦rxn is

�H◦rxn =

[∑c(�Hf (C)) + d(�Hf (D))

]−

[∑a(�Hf (A)) + b(�Hf (B))

]

HESS’S LAW OF HEAT SUMMATION

Some reactions are the sum of individual reaction steps. To sum reactions, any com-

pound that shows up on the left side of one equation and on the right side of another

equation cancels out. Hess’s law states that the sum of the �Hs for each individual

equation sum to the �Hrxn for the net equation.

2C + ��2O2 → ��2CO2 �H1��2CO2 → 2CO +��O2 �H2

sum 2C + O2 → 2CO �Hsum = �H1 + �H2

EXAMPLE: Determine the �H of the following reaction:

2S + 3O2 → 2SO3

given these reactions and their �Hs

SO2 → S + O2 �H1 = 297 kJ

2SO3 → 2SO2 + O2 �H2 = 198 kJ

SOLUTION:

➤ First, reverse the first equation and reverse the sign of �H1.

➤ Next, double the first equation and double the �H1.

➤ Then, reverse the second equation and reverse the sign of �H2.

2S + 2O2 → ��2SO2 �H1 = −594 kJ��2SO2 + O2 → 2SO3 �H2 = −198 kJ

sum 2S + 3O2 → 2SO3 �Hsum = −792 kJ

Bond Dissociation EnergyIf the equation is doubled, then the �H◦

rxn doubles as well.




and Kinetics

The �H◦f of a compound can be calculated using bond dissociation energies, found

in the following table. The formula is:

�H◦f = ∑

BE of bonds broken − ∑BE of bonds formed

where BE is the bond energy per mole of bonds.

TABLE 10-2 Dissociation Energies of Selected Single Bonds in kJ/mole

H C N O S F Cl Br IH 432C 411 346N 386 305 167O 459 358 201 142S 363 272 226F 565 485 283 190 284 155Cl 428 327 313 218 255 249 240Br 362 285 201 217 249 216 190I 295 213 201 278 208 175 149

Data from Huheey, J. E., Keiter, E. A., and Keiter, R. L., Inorganic Chemistry, 4th ed. New York:Harper Collins, 1993; pp. A21–A34.

EXAMPLE: Determine the �H◦f of HF using bond dissociation energies.

SOLUTION:

➤ First write the equation for the formation of HF from its elements:

H2 (g) + F2 (g) → 2HF (g)

➤ For H2, the BE = 432 kJ/mole.

➤ For F2, the BE = 155 kJ/mole.

➤ For HF, the BE = 565 kJ/mole.

➤ Substitute the values into the equation and solve for �H◦f :

�H◦f = [(1 mole H2)(432 kJ/mole) + (1 mole F2)(155 kJ/mole)]

− [(2 mole HF) (565 kJ/mole)] = −544 kJ

Free EnergyFree energy has the symbol G and is defined as H − T S. One can predict the spontane-

ity of a process from the sign of �G◦rxn. Free energy is useful because it eliminates the

need to worry about the surroundings.

SPONTANEOUS ENERGY

The equation used is �G◦ = �H◦ − T�S◦. When �G◦ is negative, the reaction is spon-

taneous as written. When �G◦ is positive, the reaction is spontaneous in the opposite

direction.



The maximum work for a spontaneous reaction is equivalent to �G. The free

energy change is the maximum energy available to do useful work.

THE RELATIONSHIP BETWEEN THE EQUILIBRIUMCONSTANT K AND ΔG

When not at equilibrium, �G = �G◦+RT ln Q, where Q is the reaction quotient. There-

fore, because �G = 0 at equilibrium, �G◦ = −RT ln K .

EXAMPLE: Calculate K for a reaction for which �G◦ = −13.6 kilojoules (kJ)

SOLUTION:

ln K = −13, 600 J(−8.314 J/mol K) (298.2 K)

= 5.49

K = e 5.49 = 242 large K , spontaneous reaction

ΔG AND TEMPERATURE

Because �H◦ − T�S◦ = �G◦, the signs of each variable determine the sign of �G◦. If

both �H◦ and �S◦ are positive, �G◦ is negative at high temperatures. If both �H◦ and

�S◦ are negative, �G◦ is negative at low temperatures. If �H◦ is negative and �S◦ is

positive, �G◦ is always negative. The last case is if �H◦ is positive and �S◦ is negative;

in this case, �G◦ is always positive.

EXAMPLE: Estimate the temperature above which a certain reaction becomes

spontaneous, if �H = 178.3 kJ and �S = 159 J/K.

SOLUTION:

➤ At the point between spontaneity and nonspontaneity, at equilibrium,

�G◦ = 0

➤ Plug in the values into T = �H◦/�S◦.

T = 178,300 J/159 J/K = 1121 K

Thermal ExpansionThermal expansion is a physical phenomenon in which increases in temperature can

cause substances in the solid, liquid, and gaseous states to expand.

LINEAR EXPANSION OF SOLIDS

A solid subjected to an increase in temperature �T experiences an increase in length

�L that is proportional to the original length Lo of the solid. The relation for the linear

expansion of solids is:

�L = αLo�T




and Kinetics

where the proportionality constant α is the coefficient of linear expansion and is ex-

pressed in units of inverse temperature.

EXAMPLE: A metal cylindrical rod of length L = 4.0 meters (m) is heated from

25 ◦C to 225 ◦C. Given a coefficient of linear expansion of 11 × 10−6 ◦C−1, deter-

mine the change in length following expansion.

SOLUTION: The equation needed to solve this problem is:

�L = αLo�T =(

11 × 10−6 ◦C−1) (

4.0 m) (

225 ◦C − 25 ◦C) = 8.8 mm

VOLUMETRIC EXPANSION OF LIQUIDS

A liquid subjected to an increase in temperature �T experiences an increase in volume

�V that is proportional to the original volume Vo of the liquid. The relation for the

volumetric expansion of liquids is:

�V = βVo�T

where the proportionality constant β is the coefficient of volumetric expansion and is

equal to 3α. As is the case for α, the units of β are inverse temperature.

VOLUMETRIC EXPANSION OF GASES

The volumetric expansion of gases can be summarized by two important gas laws:

Charles’s law and Boyle’s law, which can be combined in the ideal gas law.

Charles’s law states that at constant pressure P, the volume V occupied by a given

mass of gas is directly proportional to the absolute temperature T by:

V = kT orV1

T1= V2

T2

P = constant and k is a proportionality constant.

Boyle’s law states that at constant temperature, the volume V occupied by a given

mass of gas is inversely proportional to the pressure P exerted on it:

PV = k or P1V1 = P2V2

T = constant and k is a proportionality constant.

The ideal gas law combines the relations from Charles’s law and Boyle’s law to

yield:

PV = nRT orP1V1

T1= P2V2

T2

where R is the universal gas constant (R = 0.082 L atm/mol K) and n is the number of

moles.

Heats of Vaporization and FusionThe equation for heat given previously cannot be used for a phase change, because

there is no temperature difference during a phase change. The temperature remains



the same during the freezing/melting process or during the boiling/condensation pro-

cess. So for an equation that denotes a phase change, such as:

H2O (s) → H2O (l) �Hfus = 6.01 kJ/mole,

the heat change is the heat needed to convert 1 mole of solid water to 1 mole of liquid

water at 0 ◦C, and is called the heat of fusion, �Hfus.

For the boiling process,

H2O (l) → H2O (g) �Hvap = 44.6 kJ/mole

The heat change is the heat needed to convert 1 mole of liquid water to 1 mole of water

vapor at 100.0◦C, and is called the heat of vaporization, �Hvap.

The preceding can be summarized in the following table:

TABLE 10-3 Summary of Phase Transformations

Phase Descriptive Temperature Heat ofTransformation Term Point Transformation

Solid state → Liquid state Melting Melting point Heat of fusionLiquid state → Solid state Freezing Melting point Heat of fusionLiquid state → Gas state Boiling Boiling point Heat of vaporization

Gas state → Liquid state Condensation Boiling point Heat of vaporization

Phase DiagramsPhase changes occur due to a change in temperature, pressure, or both. The most com-

mon phases are solid, liquid, and gas. The names of the phase changes are as follows:

TABLE 10-4 Summary of Phase-to-Phase Changes

Solid to liquid Melting

Solid to gas Sublimation

Liquid to solid Freezing

Liquid to gas Evaporation

Gas to liquid Condensation

Gas to solid Deposition

A phase diagram shows the different phases that a substance is in over a range of

temperatures and pressures. A simple phase diagram is shown in Figure 10-4.

The lines in Figure 10-4 represent the boundaries between the phases. Each point

on a line represents a certain temperature and pressure for that phase transition. The

solid–liquid boundary line extends indefinitely, but the liquid–gas boundary ends at

the point C, which is called the critical point, and it represents the beginning of the

supercritical fluid phase. At all temperatures and pressures beyond those at point C,

the substance is in a phase that is in between liquid and gas called supercritical fluid.

Point T is called the triple point. At this pressure and temperature, all three phases

coexist in an equilibrium mixture. This means that the actual number of molecules




and Kinetics

FIGURE 10-4 A typical pressure–temperature diagram.

in each phase remains the same, but the individual molecules are changing their

phase.

The boiling and freezing points at 1 atmosphere (atm) pressure are called the nor-

mal boiling point and the normal freezing point. To determine these values, find the

temperature that corresponds to 1 atm pressure on the liquid–gas boundary and on

the solid–liquid boundary.

The density of a substance increases with pressure. Therefore, the relative densities

of the phases can be determined by following the phases as the pressure is increased.

The phase that exists at the highest pressures is the densest phase. For water, the dens-

est phase is the liquid phase. For most other substances, the solid phase is the densest

phase.

KINETICS AND EQUILIBRIUM: RATE PROCESSESIN CHEMICAL REACTIONSKinetics is the study of how fast reactions occur; this is called the reaction rate. Reac-

tion rates are dependent on the temperature at which the reaction is taking place, on

the concentrations of the reactants, and on whether a catalyst is present.

Reaction RatesReaction rates are measured experimentally. If the reactant is colored, its absorbance

can be followed. At the beginning of the reaction, the absorbance is at a maximum, and

it decreases as the reactant disappears (turns into product; see Figure 10-5).

The rate equals the change in concentration with time. The rate is fastest at the

beginning of the reaction. An instantaneous rate is the change in the concentration of

the reactant divided by the change in time, or:

�[reactant]/�t



FIGURE 10-5 Reaction rate.

when the change in time, or �t, is a very small increment of time. This is seen as the

tangent to the curve, at any point. The tangent has the steepest slope at the beginning

of the reaction (fastest rate), and the flattest slope at the end of the reaction (slowest

rate). So the reaction slows with time.

FACTORS THAT INFLUENCE REACTION RATE

The rate at which reactants become products can be manipulated via a number of

methods. By no means should you confuse these methods for obtaining more products

with the methods for shifting an equilibrium to obtain more product. Two conditions

that are needed for a reaction to occur are (1) the reactants’ molecules need to collide,

and (2) they must do so effectively.

Concentration of Reactants. In order to make more products, reactants need to be

in contact with each other. Increasing the concentration of the reactants increases the

frequency and likelihood that this process will occur. Consider the following situation.

When less A and B are present, their likelihood of collision is low:

BA

But if you double the number of A and B, you see that there are more possibilities:

B

B

A

A

Temperature. An increase in temperature will increase the average kinetic energy

of the molecules. With more kinetic energy, the molecules will have a greater velocity

according to the equation KE = 12 mv2. If the molecules collide with more kinetic

energy, their collisions will be more effective. Therefore, with an increase in kinetic

energy, the rate of reaction increases as well.

Activation Energy. Energy, no matter its quantity or its form, is needed to start a

reaction. The energy barrier that reactants need to overcome to react is called the




and Kinetics

activation energy, Ea. For an exothermic reaction, the activation energy will be small

compared to the activation energy for the reverse reaction. A good example is a stick

of dynamite. It takes only a little bit of fire (activation energy) from a match to light the

fuse, but the amount of energy released will be enormous in comparison. For an en-

dothermic reaction, the activation energy will be far greater than the activation energy

for the reverse reaction.

Physical State. The physical state of the reactants can greatly determine how long

it will take for a reaction to take place. Consider the reaction between aqueous silver

nitrate and aqueous sodium chloride. When the two reactants are mixed, the following

reaction occurs:

AgNO3 (aq) + NaCl (aq) → NaNO3 (aq) + AgCl (s)

The mixing of the two clear aqueous solutions immediately produces a white pre-

cipitate, silver chloride. The aqueous ionic solutions have their ions ready to react upon

contact.

In contrast, consider a reaction between two covalently bonded compounds such

as an alcohol and a carboxylic acid. Even though the reaction is typically catalyzed by

sulfuric acid, it needs to be refluxed for multiple hours to ensure a high yield of the

ester. In order to get the reactants to form an activated complex and start forming prod-

ucts, covalent bonds need to be broken with a sufficient amount of energy. This is a

much slower process when compared to aqueous ions.

CATALYSTS

Catalysts speed a reaction rate by lowering the activation energy of the reaction. The

catalyst may be involved in the formation of the activated complex, the high-energy

compound that is partly starting material and partly product, but it is regenerated at

the end of the reaction.

Catalysts can be in the same phase as the reactants (homogeneous) or in a differ-

ent phase (heterogeneous). They can be inorganic or organic. Biological catalysts are

called enzymes, and they catalyze every biochemical reaction in the body. Enzymes

are very specific for their starting material, which is called its substrate. They often

interact with the substrate like a key fitting into a lock. A three-dimensional site on the

substrate acts as the lock, and the enzyme fits into the site, called the active site, like a

key. Enzymes work optimally within a very narrow range of temperature and pH.

RATE LAW AND RATE CONSTANT

Kinetics studies how the reaction rate changes when the concentrations of the reac-

tants change. This rate data is used to determine the rate law for the reaction. The rate

law expresses how the rate is affected by the concentrations and by the rate constant.



For a reaction

A + 2B → products,

the rate law is written

Rate = k[A]x [B]y

where [A] is the molarity of A, [B] is the molarity of B, x is the order with respect to A,

and y is the order with respect to B.

The rate constant can be used to determine the concentration of the reactant at

any point. For a first-order reaction (overall), the equation is:

ln[A]t = ln[A]0 − kt

where ln means natural log, [A]0 is the concentration of A at the beginning of the reac-

tion (t = 0), and [A]t is the concentration of A at time t; k is the rate constant, and t is

the time.

EXAMPLE: For a first-order reaction

A → product,

the rate constant is 3.02 × 10−3 s−1. The concentration of A initially is 0.025 M.

SOLUTION:

1. Calculate the [A] after 172 s.

2. Calculate the time for the [A] to reach 1.13 × 10−3 M.

3. Calculate the time needed for 98% of the [A] to react.

➤ Plug in [A]0 = 0.0250 M, t = 172 s, k = 3.02 × 10−3 s−1

ln [A]t = ln 0.0250 − (3.02 × 10−3 s−1)(172 s)

= −3.69 − 0.519

= −4.21

[A]t = e−4.21 = 0.015 M

➤ln[A]t

ln[A]0= −kt

ln0.015

0.0250= −(3.02 × 10−3 s−1) · t

−0.511 = −3.02 × 10−3 s−1 · t

t = 0.511

3.02 × 10−3 s−1= 169 s




and Kinetics

➤ln[A]t

ln[A]0= −kt

ln2%

100%= −(3.02 × 10−3 s−1) · t

t = 1294 s

REACTION ORDER

Rates usually relate to the concentration of a reactant in one of three ways: by not

changing, by changing proportionately to the change in concentration, or by changing

as the square of the change of the concentration.

➤ As the concentration of A is changed, the rate does not change (see Figure 10-6a).

The rate is unaffected; the order with respect to A is 0.

➤ As the concentration of A is changed, the rate changes proportionally (see Fig-

ure 10-6b). If [A] is doubled, the rate doubles, and so on. The order with respect to

A is 1.

➤ As the concentration of A is changed, the rate changes as the square of the

change in the concentration (see Figure 10-6c). If [A] is doubled, the rate quadru-

ples, for instance. The order with respect to A is 2.

FIGURE 10-6 Relation of rate to concentration of reactant.

The orders of each reactant are thus determined by observing how the rate is

affected by the change in the concentration of each reactant. Only one order can be

determined at a time. All other concentrations must remain constant; only the con-

centration of the reactant being studied is changed. The rates are shown under various

conditions; these are used to determine the rate law and the rate constant, k, is then

calculated.

EXAMPLE: For the reaction

A + B + C → products,



determine the rate law and calculate the rate constant using the following data:

[A] [B] [C] RateRun (M) (M) (M) (M/s)

1 0.003 0.001 1.0 1.8 × 104

2 0.003 0.002 1.0 3.6 × 104

3 0.006 0.002 1.0 7.2 × 104

4 0.003 0.001 2.0 7.2 × 104

SOLUTION:

➤ From run 1 to 2, the [B] is doubled. The rate doubles, so the order of B is 1.

(The rate is directly proportional to the concentration of B.)

➤ From run 2 to 3, the [A] is doubled. The rate doubles, so the order of A is 1.

(The rate is directly proportional to the concentration of A.)

➤ From run 1 to 4, the [C] is doubled. The rate quadruples, so the order of C is 2.

(The rate is proportional to the square of the concentration of C.)

➤ So the rate law is:

rate = k[A] [B] [C]2

➤ Using the data from run 1, the rate constant k is calculated:

1.8 × 10−4 = k[0.003] [0.001] [1]2

k = 60 M−3s−1

Rate-Determining StepNot all reactions take place in a single step. Instead, many reactions have multiple

steps, each with its own activation energy. These individual steps are called elemen-

tary steps and, when added together, produce the overall reaction. The step with the

greatest energy barrier to overcome will be the slowest step. It is considered to be the

rate-determining step. For example, when you consider the overall reaction:

2NO (g) + Cl2 (g) → 2NOCl (g),

there are actually two elementary steps:

Step 1: NO + Cl2 → NOCl2

Step 2: NOCl2 + NO → 2NOCl

To actually know the rate-determining step, you must know the observed rate law.

For this particular reaction, the slower step is the second elementary step.

How would you know if the reaction takes place in one step? According to the over-

all reaction, two molecules of nitric oxide gas, NO(g), must make contact and react with

chlorine gas to form the product. It is highly unlikely that a termolecular reaction will




and Kinetics

take place. Instead, you see that there are bimolecular reactions taking place to form

an intermediate, which is consumed in the next step. If one substance were to form a

product or intermediate, that type of reaction would be considered to be unimolecular.

Activated Complex or Transition StateActivation energy is needed to provide reactants with enough energy to produce an

intermediate between the reactants and products. Once this energy has been achieved,

the reaction then reaches a transition state in which the reactants start to form the

products. This transition state is also called the activated complex. The complex

formed has its own potential energy called the potential energy of the activated com-

plex. This point of a potential energy diagram will be a maximum level, meaning that

the complex is less stable and that the next steps of the reaction will be exothermic. This

complex is usually a short-lived substance. Once the complex is formed, the products

begin to form immediately so as to become a lower-energy, more stable substance.

Interpretation of Energy ProfilesIn order to track the energy changes over the course of a reaction, an energy profile (or

reaction coordinate diagram) is used. A typical energy profile looks like Figure 10-7.

A

Pote

ntia

l ene

rgy

Reaction coordinate

BC

D

E

FIGURE 10-7 Typical energy profile.

Figure 10-7 has the following features:

A: The potential energy of the reactants

B: The activation energy of the forward reaction

C: The activation energy of the reverse reaction

D: The heat of reaction, �H

E: The potential energy of the products

An example of a multistep energy profile is demonstrated in Figure 10-8. The first

energy barrier, requiring more activation energy, would be the rate-determining step

for this sample reaction.



Pote

ntia

l ene

rgy

Reaction pathway

FIGURE 10-8 Multistep energy profile.

ARRHENIUS EQUATION

If the energy of activation, Ea, for a reaction is known (this is the minimum energy

needed for a reaction to occur), and the rate constant for one temperature is known,

the rate constant at any other temperature can be calculated using the Arrhenius

equation:

lnk2

k1= Ea

R(1/T1 − 1/T2)

where k1 is the rate constant at temperature T1, k2 is the rate constant at temperature

T2, T1 and T2 are in Kelvin, R = 8.314 joules (J)/mole K, and Ea is the activation energy.

EXAMPLE: For a certain first-order reaction, the rate constant at 190 ◦C is 2.61

× 10−3 and the rate constant at 250 ◦C is 3.02 × 10−3. Calculate the energy of

activation of this reaction.

SOLUTION:

k1 = 2.61 × 10−3 at T1 = 463.2 K

k2 = 3.02 × 10−3 at T2 = 523.2 K

So

ln3.02 × 10−3

2.61 × 10−3= Ea

8.314

( 1463.2

− 1523.2

)

ln 1.16 = Ea(2.48 × 10−4)

8.314

0.148(8.314)

2.48 × 10−4= Ea

5.0 × 103 J/mole = Ea

Kinetic Control Versus Thermodynamic Controlof a ReactionSometimes a reaction has competing pathways that can lead to different products,

even though the reactants are exactly the same. One pathway may require a lower




and Kinetics

activation energy, or one may form a more stable product than another. The reac-

tion that is governed by a lower activation energy will produce a kinetically-controlled

product. A reaction that is favored because it forms a more stable product does so

by forming a thermodynamically-favored product. The energy profile shown in Fig-

ure 10-9 summarizes these two competing possibilities.

Energy ofreactants

EA

Progress of reaction

Ener

gy

Product AProduct B

FIGURE 10-9 Kinetic control versus thermodynamic control.

Product A was the kinetically-favored (lower activation energy) product, while Prod-

uct B was formed via a thermodynamically-favored pathway (more stable product).

Equilibrium in Reversible Chemical ReactionsEquilibrium occurs with reactions that are reversible when the forward reaction rate

equals the reverse reaction rate. Using various formulas, the rate at which a reaction

occurs before reaching equilibrium and the concentrations of the products can be

determined.

For a reversible reaction, the starting material (on the left) is converted into the

product (on the right) at a rate equal to k[A]x:

A � B

At a certain point in the progress of the reaction, some of the B that has formed

begins to convert to A at a rate equal to k−1[B]y. There comes a point when the forward

rate and the reverse rate are equal, and the concentrations of A and B no longer change,

although each reaction is still proceeding. This is called the equilibrium point. The net

numbers of A and B present in the reaction flask remain the same. It appears that the

reaction has stopped.



EXAMPLE: If one begins with 100 molecules of A, A begins immediately to turn

into B. When enough B’s are present, they start to turn back into A. At the equili-

brium point, there are, for example, 60 A’s and 40 B’s. The reactions, both forward

and reverse, are still proceeding, but the reaction rates are equal, so there remain

at all times 60 A’s and 40 B’s.

LAW OF MASS ACTION

For a reaction at equilibrium, there is a method in which chemists express the concen-

tration of products formed to the reactants remaining. This is called the law of mass

action. For example, the following reaction is at equilibrium: wW + xX ←→ yY + zZ.

The law of mass action expresses the ratio of products formed to reactants remaining

where the concentration of each substance is raised to a power that is equal to their

coefficient. Using the preceding reaction at equilibrium, the law of mass action would

look like:

Keq = [Y ]y[Z]z/[W]w[X]x

A good mnemonic device to remember is, “Products over reactants; coefficients

become powers.”

The phases of the substances involved in the equilibrium are important to note

as well. The law of mass action will include only aqueous substances (aq) or gaseous

substances (g). It will not include solids (s) or liquids (l). When including an aqueous

substance, the concentration should be noted in molarity, M. When gases are involved,

the partial pressure of each gas is used. Applying these rules to a purely hypothetical

reaction at equilibrium: 2W (aq) + 3X (s) ←→ 4Y (l) + 5Z (g) the law of mass action

would result in:

Keq = [Z]5/[W]2

EQUILIBRIUM CONSTANT

The equilibrium constant K describes the extent to which the forward reaction pro-

ceeds before reaching the equilibrium point. Is there a lot of A left over, or just a little?

K is a constant value at a constant temperature; it does change with temperature.

If K is large (>1), there is mostly product and very little starting material left at the

equilibrium point. This is a product-favored reaction.

If K is small (<10−4), there is mostly starting material left over, and very little prod-

uct formed at the equilibrium point. This is a reactant-favored reaction.

With a medium K (between 1 and 10−4), a lot of product is formed, but there are

still substantial quantities of starting material left.




and Kinetics

Factors Affecting Equilibrium Constant—Equilibrium Expression. The equi-

librium constant K depends on temperature and is related to the amounts of starting

material and product in the following manner:

a A + b B � c C Kc = [C]c

[A]a[B]b

where a, b, and c are the molar coefficients and [ ] = mole per liter (mol/L)

EXAMPLE:

N2 + 3 H2 � 2NH3

SOLUTION:

Kp = (PNH3)2

(PN2)(PH2

)3in terms of pressure

Only gases (with pressures) and species in solution (with concentrations) affect the

equilibrium. Solids and liquids do not appear in the equilibrium expression.

EXAMPLE:

C (s) + 2Cl2 (g) � CCl4 (g)

SOLUTION:

Kp = (PCCl4)

(PCl2)2

Given the K and the initial conditions (concentrations or pressures), you can cal-

culate the equilibrium conditions of the products.

If the K is <10−4, the reaction does not proceed far before equilibrium. Very little

product is formed, and the amount of starting material has changed very little. You can

neglect the change in the starting material because it is such a small amount.

EXAMPLE: If you start with 0.240 mole SO3 in a 3.00-L container, calculate the

equilibrium concentrations of SO2 and O2 in the reaction:

2SO3 � 2SO2 + O2 K = 9.6 × 10−8

SOLUTION:

➤ The balanced chemical equation is used to set up an ICE (Initial, Change,

Equilibrium) table



TABLE 10-5 ICE Table for a Balanced Chemical Equation

2SO3 � 2SO2 + O2Initial 0.080 M 0 0 Initial amounts in M or pressure

� −2x +2x +x For every 2 moles of SO3 that react,2 moles of SO2 and 1 mole of O2 areformed.

Equilibrium 0.080 M 2x x Add the values in the column to get theequilibrium value; you can neglect thechange to the SO2; these values go intothe K expression.

➤ Write the K expression:

K = [SO2]2 [O2]

[SO3]2

➤ Plug in the equilibrium values and solve for x:

9.6 × 10−8 = [2x]2 [x]

[0.080]2

➤ x = 5.4 × 10−4

➤ So the equilibrium concentrations of SO2 and O2 are:

[SO2] = 2x = 1.1 × 10−3 M

[O2] = x = 5.4 × 10−4 M

You can calculate the K if given the initial conditions (concentrations or pressures)

and either the percentage of starting material that reacts or one of the actual equili-

brium values.

EXAMPLE: You have 0.40 atm of CH4 and 0.65 atm of H2S initially. Calculate the

K of the following reaction if 25% of the CH4 remains at the equilibrium point.

SOLUTION:

CH4 (g) + 2H2S (g) � CS2 (g) + 4H2 (g)Initial 0.40 0.65 0 0

� −x −2x +x +4x

At equilibrium 0.10 0.05 0.30 1.20

➤ The numbers on the bottom line were determined using the information given

in the problem. If you start with 0.40 atm of CH4 and 25% remains at the equi-

librium point, then (0.40 atm) (0.25) = 0.10 atm is what remains of the CH4 at

equilibrium.

➤ Then using the equation 0.40 − x = 0.10, you get x = 0.30 atm.

➤ Plugging in this value for x on the change line and adding the columns, the




and Kinetics

other equilibrium values are obtained.

➤ Now you write the K expression.

K = (PCS2) (PH2

)4

(PCH4) (PH2S)2

➤ Plug in the bottom line values to calculate K .

K = (0.30) (1.20)4

(0.10) (0.05)2= 2.5 × 103

LE CHATELIER’S PRINCIPLE

Le Chatelier’s principle states that if you subject a system that is at equilibrium, to

some change in conditions, the equilibrium shifts so as to counteract the change. For

example, note the following reaction:

2FeCl3 (s) + 3H2O (g) � Fe2O3 (s) + 6HCl (g) endothermic

TABLE 10-6 Summary of Changes in Reactions in Equilibrium

Direction ofChange Counteraction Equilibrium Shift

Add some water vapor Remove the water vapor −→Remove some HCl Make more HCl −→Remove some Fe2O3 No counteraction No change

Add some HCl Remove some HCl ←−Increase the volume Increase the pressure −→

(decrease the pressure) (make more gas molecules)

Decrease the volume Decrease the pressure ←−(increase the pressure) (remove gas molecules)

Heat the reaction Remove the heat −→Cool the reaction Increase the heat content ←−

CHAPTER 10 Principles of Chemical Thermodynamics and · PDF fileenergy transfer, in which...

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